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walters
 2 years ago
Best ResponseYou've already chosen the best response.0jjj let the mapping \[\tau _{ab}\] for a,b element R ,maps the reals by the ruel \[\tau _{ab}:x \rightarrow ax+b.Let G={\tau _{ab}:a \neq 0}\] Determine whether or not G is an abelian group under the composition of mappings

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.4Well, let \(\tau_{ab},\tau_{cd}\in G\). Then \[\tau_{ab}\circ\tau_{cd}(x)=\tau_{ab}(cx+d)=acx+ad+b.\]Also, \[\tau_{cd}\circ\tau_{ab}(x)=\tau_{cd}(ax+b)=cax+bc+d.\]Since \(acx+ad+b\neq acx+bc+d\), it must be that \(G\) is not abelian.

walters
 2 years ago
Best ResponseYou've already chosen the best response.0am i not suppose to check whether is a group using the condition before i can conclude that is an abelian group

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.4It just asked if it was an abelian group. Since the elements don't commute, it can't be an abelian group even if it is a group. However, if we want to check it's a group, we can do that as well. Have you shown any of the axioms for proving it's a group yet?

walters
 2 years ago
Best ResponseYou've already chosen the best response.0i am failing to show the axioms

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.4First, we can see that \(G\) is closed under composition. \[\tau_{ab}\circ\tau_{cd}(x)=\tau_{ab}(cx+d)=acx+ad+b=\tau_{(ac)(ad+b)}.\]We also have the identity \(\tau_{10}\). Now we just have to check associativity and inverses.

walters
 2 years ago
Best ResponseYou've already chosen the best response.0why do u chose to use dw:1360875075213:dw

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.4"\(\circ\)" is a relatively common notation to denote the composition of two functions. You can also use \(\tau_{ab}(\tau_{cd}(x))\) is you prefer.

walters
 2 years ago
Best ResponseYou've already chosen the best response.0can u also use * if u wnt

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.4For inverses, we want \[\tau_{ab}\circ\tau_{cd}(x)=acx+ad+b=x\]for some \(c,d\in\mathbb{R}\). We can see that \(c=a^{1}\) and \(d=a^{1}(b)\) work for this direction. For the other, \[\tau_{cd}\circ\tau_{ab}(x)=cax+bc+d=a^{1}ax+ba^{1}a^{1}b=x,\]so we do indeed have inverses.

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.4If you make it clear your group function is composition of functions, you can use * as well.

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.4As for associativity, I'm just going to use the fact that composition of functions is always associative. If you have not heard of that yet, we can still go over how to prove this particular function is associative.

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.4let \(\tau_{ab},\tau_{cd},\tau_{ef}\in G\). Then\[\tau_{ab}\circ(\tau_{cd}\circ\tau_{ef})(x)=\tau_{ab}=\tau_{ab}(cex+cf+d)=acex+acf+ad+b\]and\[(\tau_{ab}\circ\tau_{cd})\circ\tau_{ef}(x)=(\tau_{(ac)(ad+b)})(ex+f)=acex+acf+ad+b\]Since these are equal, it's associative, and we have a group.

walters
 2 years ago
Best ResponseYou've already chosen the best response.0k it is a group but not an abelian group

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.4\[\tau_{10}(x)=1x+0=x\]
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